Question

1 cm on the main scale of Vernier callipers is divided into 10 equal parts, if 10 vernier scales coincide with 8 small divisions of main scale, then the least count of the calliper is
(A) $0.01cm$
(B) $0.02cm$
(C) $0.05cm$
(D) $0.005cm$

Answer 1

The least count is directly related to the width of each division of the main scale. 10 divisions of the Vernier scale coincides with 8 divisions of the main scale, the least count is proportional to ratio of the difference to the number of divisions of the Vernier scale.

Formula used: In this solution we will be using the following formulae;
$MR = \dfrac{L}{N}$ where $MR$ signifies minimum reading of the main scale, $L$ is the length of a section of the main scale and $N$ is the number of division in that section.
$LC = \dfrac{{MR}}{n}$ where $LC$ is the least count (without coincidence error)$n$ is the number of divisions on the Vernier scale

Complete Step-by-Step solution:
To calculate the least count, we find the smallest reading on the main scale. This is equal to the length of a particular section of the Vernier calliper’s main scale divided by the number of divisions of that section. In the question, we are told a 1 cm section is divided into 10 divisions, i.e.
$\Rightarrow$ $MR = \dfrac{L}{N}$ where $MR$ signifies minimum reading of the main scale, $L$ is the length of a section of the main scale and $N$ is the number of division in that section.
Hence, we have
$\Rightarrow$ $MR = \dfrac{{1cm}}{{10}} = 0.1cm$
Least count can be defined as
$\Rightarrow$ $LC = \dfrac{{MR}}{n}$ where $n$ is the number of divisions on the Vernier scale
$\Rightarrow$ $LC = \dfrac{{0.1cm}}{{10}} = 0.01cm$
Nonetheless, the least count will be increased due to the “coincidence error”. Only eight divisions coincide with 10 divisions, hence, coincidence error is $10 - 8 = 2$
Hence, true $LC$ would be
$LC = 0.01cm \times 2 = 0.02cm$

Hence, the correct option is B,

Note: Alternatively, we can simply use the relation
$LC = MR - M{r_v}$ where $M{r_v}$ is the minimum reading of the Vernier scale.
Since only 8 coincides, then only 8 can be read, then $M{r_v}$ would be
$M{r_v} = \dfrac{{8mm}}{{10}} = 0.8mm = 0.08cm$
Hence,
$LC = MR - M{r_v} = 0.1cm - 0.08cm = 0.02cm$